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never [62]
3 years ago
15

A bolt is to be made 2.0 cm long with a tolerance of 4%. Set up an equation to solve for the shortest and longest bolt length. S

how your original equation and the process used to find the shortest and longest bolt length.
Mathematics
1 answer:
pickupchik [31]3 years ago
5 0

Answer:

The equations are:

\text{shortest} = \text{length} - \frac{\text{tolerance}}{100}*\text{length}\\\text{shortest} = 2 - \frac{4}{100}*2 = 1.92\\\text{longest} = \text{length} + \frac{\text{tolerance}}{100}*\text{length}\\\text{longest} = 2 + \frac{4}{100}*2 = 2.08\\

Step-by-step explanation:

The ideal length of the bolt is 2.0 cm, since it has a tolerance of 4%, this means that length values between 2 + \frac{4}{100}*2 and 2 - \frac{4}{100}*2 are acceptable, because by those expressions the length of the bolt will be 4% higher or lower than it's original length. With this in mind let's find it's values:

\text{shortest} = \text{length} - \frac{\text{tolerance}}{100}*\text{length}\\\text{shortest} = 2 - \frac{4}{100}*2 = 1.92\\\text{longest} = \text{length} + \frac{\text{tolerance}}{100}*\text{length}\\\text{longest} = 2 + \frac{4}{100}*2 = 2.08\\

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Will mark brainliest if correct answer only <br><br><br><br><br>All,some,or no ​
NeTakaya

Answer:

some

Step-by-step explanation:

8 0
3 years ago
Consider the integral Integral from 0 to 1 e Superscript 6 x Baseline dx with nequals 25 . a. Find the trapezoid rule approximat
photoshop1234 [79]

Answer:

a.

With n = 25, \int_{0}^{1}e^{6 x}\ dx \approx 67.3930999748549

With n = 50, \int_{0}^{1}e^{6 x}\ dx \approx 67.1519320308594

b. \int_{0}^{1}e^{6 x}\ dx \approx 67.0715427161943

c.

The absolute error in the trapezoid rule is 0.08047

The absolute error in the Simpson's rule is 0.00008

Step-by-step explanation:

a. To approximate the integral \int_{0}^{1}e^{6 x}\ dx using n = 25 with the trapezoid rule you must:

The trapezoidal rule states that

\int_{a}^{b}f(x)dx\approx\frac{\Delta{x}}{2}\left(f(x_0)+2f(x_1)+2f(x_2)+...+2f(x_{n-1})+f(x_n)\right)

where \Delta{x}=\frac{b-a}{n}

We have that a = 0, b = 1, n = 25.

Therefore,

\Delta{x}=\frac{1-0}{25}=\frac{1}{25}

We need to divide the interval [0,1] into n = 25 sub-intervals of length \Delta{x}=\frac{1}{25}, with the following endpoints:

a=0, \frac{1}{25}, \frac{2}{25},...,\frac{23}{25}, \frac{24}{25}, 1=b

Now, we just evaluate the function at these endpoints:

f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1

2f\left(x_{1}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281

2f\left(x_{2}\right)=2f\left(\frac{2}{25}\right)=2 e^{\frac{12}{25}}=3.23214880438579

...

2f\left(x_{24}\right)=2f\left(\frac{24}{25}\right)=2 e^{\frac{144}{25}}=634.696657835701

f\left(x_{25}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735

Applying the trapezoid rule formula we get

\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{50}(1+2.54249830064281+3.23214880438579+...+634.696657835701+403.428793492735)\approx 67.3930999748549

  • To approximate the integral \int_{0}^{1}e^{6 x}\ dx using n = 50 with the trapezoid rule you must:

We have that a = 0, b = 1, n = 50.

Therefore,

\Delta{x}=\frac{1-0}{50}=\frac{1}{50}

We need to divide the interval [0,1] into n = 50 sub-intervals of length \Delta{x}=\frac{1}{50}, with the following endpoints:

a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b

Now, we just evaluate the function at these endpoints:

f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1

2f\left(x_{1}\right)=2f\left(\frac{1}{50}\right)=2 e^{\frac{3}{25}}=2.25499370315875

2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281

...

2f\left(x_{49}\right)=2f\left(\frac{49}{50}\right)=2 e^{\frac{147}{25}}=715.618483417705

f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735

Applying the trapezoid rule formula we get

\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{100}(1+2.25499370315875+2.54249830064281+...+715.618483417705+403.428793492735) \approx 67.1519320308594

b. To approximate the integral \int_{0}^{1}e^{6 x}\ dx using 2n with the Simpson's rule you must:

The Simpson's rule states that

\int_{a}^{b}f(x)dx\approx \\\frac{\Delta{x}}{3}\left(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)\right)

where \Delta{x}=\frac{b-a}{n}

We have that a = 0, b = 1, n = 50

Therefore,

\Delta{x}=\frac{1-0}{50}=\frac{1}{50}

We need to divide the interval [0,1] into n = 50 sub-intervals of length \Delta{x}=\frac{1}{50}, with the following endpoints:

a=0, \frac{1}{50}, \frac{1}{25},...,\frac{24}{25}, \frac{49}{50}, 1=b

Now, we just evaluate the function at these endpoints:

f\left(x_{0}\right)=f(a)=f\left(0\right)=1=1

4f\left(x_{1}\right)=4f\left(\frac{1}{50}\right)=4 e^{\frac{3}{25}}=4.5099874063175

2f\left(x_{2}\right)=2f\left(\frac{1}{25}\right)=2 e^{\frac{6}{25}}=2.54249830064281

...

4f\left(x_{49}\right)=4f\left(\frac{49}{50}\right)=4 e^{\frac{147}{25}}=1431.23696683541

f\left(x_{50}\right)=f(b)=f\left(1\right)=e^{6}=403.428793492735

Applying the Simpson's rule formula we get

\int_{0}^{1}e^{6 x}\ dx \approx \frac{1}{150}(1+4.5099874063175+2.54249830064281+...+1431.23696683541+403.428793492735) \approx 67.0715427161943

c. If B is our estimate of some quantity having an actual value of A, then the absolute error is given by |A-B|

The absolute error in the trapezoid rule is

The calculated value is

\int _0^1e^{6\:x}\:dx=\frac{e^6-1}{6} \approx 67.0714655821225

and our estimate is 67.1519320308594

Thus, the absolute error is given by

|67.0714655821225-67.1519320308594|=0.08047

The absolute error in the Simpson's rule is

|67.0714655821225-67.0715427161943|=0.00008

6 0
3 years ago
A quadratic function y=f(x) is plotted on a graph and the vertex of the resulting parabola is (−5,−3). What is the vertex of the
Licemer1 [7]

Answer:

(-5 , 6)

Step-by-step explanation:

f(x) = (x - (-5))² - 3     ... vertex (h , k) : (-5 , -3)     y = (x-h)² + k

g(x) = -f(x) + 3 = -((x+5)² - 3) + 3 = -(x + 5)² + 6

vertex of g(x) is : (-5 , 6)

4 0
3 years ago
Mind if I have some assistance.
pentagon [3]
The answer is -7 because :

a=1, b=1, c=2
Discriminant = b^2 - 4ac
= 1^2 - 4*1*2
= -7
3 0
3 years ago
Write and solve an equation to find the unknown side length x (in inches). Perimeter =24.2 in. Sides of the shape are 8.3, 8.3,
Anastasy [175]

Answer:

Perimeter of shape,which is a convex polygon i.e quadrilateral = 24.2 in

Sides are 8.3 in, 8.3 in, 3.8 in ,and length of fourth side be x in.

As , Perimeter = Sum of all sides

                 → 24.2 = 8.3 + 8.3 + 3.8 + x

                 → 24.2 = 16.6 +3.8 + x

                 → 24.2 = 20 .4 + x

Keeping like terms on one side, of equation

            → 24.2 - 20.4 = x

            → 3.8 = x

            → x = 3.8

So, length of fourth side = 3.8 in

The given shape is definately either a Parallelogram or a kite.



7 0
3 years ago
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